P2 Trig! ARGHHH Watch

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kikzen
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#21
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#21
(Original post by lgs98jonee)
hang on...ur idea of using cosP -cosQ=0 works on the question doesnt it? I used that method and got all the solutions...or am i missing sumthing here
well i thought so, but Squishy seems to think otherwise?
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kimoni
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#22
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#22
(Original post by lgs98jonee)
hang on...ur idea of using cosP -cosQ=0 works on the question doesnt it? I used that method and got all the solutions...or am i missing sumthing here
works for me
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Squishy
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#23
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#23
(Original post by kikzen)
bleh.. solve it then
I was referring to your madness.
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kikzen
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#24
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#24
(Original post by Squishy)
I was referring to your madness.
oh lol heh, great sig tho
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It'sPhil...
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#25
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#25
(Original post by Katie Heskins)
Theres nothing wrong with assuming the result you're trying to prove providing you don't do anything invalid, therefore lgs98jonee's is fine because he hasn't made any errors which may have lead to an invalid proof.
This is only true if you use <= or <=> implication signs. In these proofs you are tring to prove a statement, call it Z. You start from a known statement, A and follow a logical argument A => B => C => ... => Z. If you assume Z true and work backwards to A showing that Z => Y => ... => A, This is not the same as the first unless you can show that Z <=> Y <=> ... <=> A.
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lgs98jonee
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#26
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#26
(Original post by It'sPhil...)
This is only true if you use <= or <=> implication signs. In these proofs you are tring to prove a statement, call it Z. You start from a known statement, A and follow a logical argument A => B => C => ... => Z. If you assume Z true and work backwards to A showing that Z => Y => ... => A, This is not the same as the first unless you can show that Z <=> Y <=> ... <=> A.
i didnt work backwords though..

and also when proving a trig identity u can work both ways becouse it is tautologous???
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It'sPhil...
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#27
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#27
(Original post by lgs98jonee)
i didnt work backwords though..

and also when proving a trig identity u can work both ways becouse it is tautologous???
Hehe ok, but identities are differne since LHS = RHS is one statement, you could assume it is true and show that (LHS = RHS) => ... => (1 = 1) or (0 = 0) or anything like that. This is invalid as a proof, however (0 = 0) => ... => (LHS = RHS) is ok. But the standard is to show that RHS = ... = LHS or vice versa
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